Analogue and Digital Electronics: Power Converter Design Assignment

This assignment solution addresses various aspects of power converter design in electrical engineering. It begins with the analysis of a two-transistor regulator, deriving the formula for output voltage and calculating component values. It then progresses to the design of different power converter topologies, including buck, boost, and synchronous converters, to meet specific requirements such as input voltage ranges, output voltage, load current, switching frequency, and ripple voltage. The solution includes detailed calculations for inductor and capacitor values, duty cycle determination, and analysis of circuit behavior. Furthermore, it differentiates between synchronous and typical buck converters, discusses the advantages of synchronous converters, and explores the impact of input capacitors. The assignment covers designing circuits, calculating component values, and comparing the efficiency of different regulator types.

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Student
Instructor
Analogue and digital electronics
Date

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QUESTION 1
a. Deriving the formula for Vout of a simple two transitor regulator.
Relationship between feedback voltage and output voltage
V F=V R 3= R3
{ R3 + R4 } V out (1)
Considering R3, ZD1 and Q1 loop, and applying KVL
V F−V BE 2−V VDZ 1=0
Implying that
V F=V BE 2+V VDZ 1 (2)
Replacing equation 2 into 1
V BE 2 +V VDZ 1= R3
{ R3 + R4 } V out (3)
Rearranging the equation
V out = { R3+R4 }
R3
{ V BE 2 +V VDZ1 }
b. Calculating values of the components
Finding current through R3

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I R 3 =V F
R3
But V F=V BE 2+V VDZ1=0.7+4.1 V =4.8 V
Therefore,
I R 3 = 4.8
4700 =1.021mA
Current through R2 is approximately equal to current through R3
c. Designing the power circuit of a standard power converter that would be an acceptable
replacement for the circuit.
At this high switching frequency of 150kHz, buck converter is the most suitable since it is a
voltage step down DC-DC converter
The voltage ripple allowable is 15 mV on the output. Taking ripple current as 0.1A, Inductor and
capacitor values can be estimated as follow:
Inductor
Vin: 15V and Vout: 6V
L= V out × (V ¿max
−V out )
∆ I l × f s ×V ¿max
L= { 6V × ( 15−6 ) V }
{ 0.1 A ×15 0,000 Hz× 15 V }
L=0.2 4 mH
The value of capacitor needed is given by.

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Cout= ∆ I L
8 × f s ×∆ V OUT
Cout= 0.1 A
{ 6 ×15 0,000 Hz ×0.015 V }
Cout=7.41 μF
QUESTION 2
Designing a high efficiency dc dc power converter with specifications below
Nominal load current: 5A
Nominal output voltage: 8V
Input voltage min 12V and max 18V.
Current ripple: 0.1A max
Switching frequency : 30kHz
Output ripple voltage: 20mV
a. Designing a suitable power circuit topology to meet the above requirements.
The suitable power converter is buck converter since the output voltage is a stepped down version
of the input voltage and the switching frequency if very high.
In this design, estimation of the inductor value depends on the ripple current, switching frequency,
input voltage and output voltage.
L= V out × (V ¿max
−V out )
∆ I l × f s ×V ¿max

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L= {8V × ( 18−8 ) V }
{ 0.1 A ×30,000 Hz ×18 V }
L=1.48 mH
The value of capacitor depends of the ripple current, output ripple voltage and switching frequency.
Cout= ∆ I L
8 × f s ×∆ V OUT
Cout= 0.1 A
{ 8 ×30,000 Hz ×0.02 V }
Cout=20.8 μF
b. Calculating the minimum and maximum duty cycle
V omin
=V o−∆ V o=8 V −20 mV
V omin
=7.98 V
Maximum output voltage
V omax
=V o +∆ V o =8 V +20 mV
V omin
=8.02VV
Maximum duty cycle of the design
Dmzx =V omax
V ¿min
=8.02
12 ≈ 0.68
Minimum duty cycle of the design
Dmin= V omix
V ¿man
= 7.98
18 ≈ 0.4 4
c. Calculating the maximum input current
Imax=I out × Dmax
I max=5 A × 0.68=3.4 A
QUESTION 3
The synchronous converter has the following requirements.
Vin: 350V

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Vout: 75V regulated
Load Power: 200W
Max. output ripple: 30mV
Switching frequency: 750kHz
a. Calculating L1 and C1.
Ripple current is assumed to be 0.1A
Modelling Inductor by assuming inductor ripple current as 0.1A
L= V out × (V ¿max
−V out )
∆ I l × f s ×V ¿max
L= {75V × ( 350−75 ) V }
{ 0.1 A ×750 , 000 Hz × 350 V }
L=0.786 μ H
Modelling the capacitor.
Cout= ∆ I L
8 × f s ×∆ V OUT
Cout= 0.1 A
{ 8 ×750 , 000 Hz× 0. 0 3V }
Cout=0. 5 56 n F
b. Defining synchronous converter and differentiating with a typical buck converter
Synchronous converter is a DC DC power converter that uses a combination of two Mosfet
switched that operates 1800 out of phase to step down DC voltage while stepping up DC
current.
Different between synchronous buck converter and typical buck converter.
Synchronous buck converter has two MOSFET switches while typical buck converter
uses a combination of one switch and a diode.
QUESTION 4.
Designing a boost converter as per the specifications below.
Vout: 75V
Vin: 30V regulated

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Load power: 200W
Switching frequency: 750 kHz.
Finding the output load resistance
RL= V out
2
Pout
= 752
200 =28.125 Ω
Finding the maximum duty cycle of the boost converter
V out = {V ¿ }
{1− D }
Implying that
D= {V ¿−V out }
−V o
= 30 V −75 V
−75 V =0.6
Finding the value of inductor;
Lout= { ( 1−D ) 2 DR }
2 f
Lout= { (1−0.6 )2 ×0.6 × 28.125Ω }
{2 ×750,000 Hz }
Lout=1.8 mH
Finding the value of the capacitor
C= { DV out }
{ ∆ VRf }
Assuming ∆ V =20 mV
C= { 0.6 ×75 }
{ 0.02V ×28.125 Ω ×750,000 Hz }
C=1.067 μF

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QUESTION 5
Designing a high efficiency d.c to d.c power converter with the given specifications
Input voltage: 10-15V
Output nominal voltage: 8 V
Nominal load current: 4A
Inductor current ripples: 0.1A
Switching frequency: 30 kHz
Output voltage ripple: 20 mV
d. Defining a suitable power circuit topology to meet the above specifications.
Since the converter is a voltage-step down converter, DC-DC buck converter is the most suitable.

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e. Defining the minimum and maximum duty cycles assuming that the control circuit keeps the
output voltage constant at the nominal value.
The buck converter is to be operated in Continuous Current Mode (CCM).
Minimum output voltage [1].
V omin
=V o−∆ V o=8 V −20 mV
V omin
=7.98 V
Maximum output voltage
V omax
=V o +∆ V o =8 V +20 mV
V omin
=8.02VV
Maximum duty cycle of the design
Dmzx =V omax
V ¿min
=8.02
10 ≈ 0.8
Minimum duty cycle of the design
Dmix = V omix
V ¿man
=7.98
15 ≈ 0.5
f. Finding the maximum input current (Assuming the load current is constant at the nominal value)
I max=I out × Dmax
I max=4 A × 0.8=3.2 A

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g. Designing a suitable converter power circuit using a Mosfet switch, showing all calculation of
inductor and capacitor values and drawing.
Estimation of the inductor value is determined by the expression below
L= V out × (V ¿max
−V out )
∆ I l × f s ×V ¿max
L= {8V × ( 15−8 ) V }
{ 0.1 A ×30,000 Hz ×15 V }
L=1.244 mH
The value of capacitor needed for the buck converter design is given by.
Cout= ∆ I L
8 × f s ×∆ V OUT
Cout= 0.1 A
{ 8 ×30,000 Hz ×0.02 V }
Cout=20.8 μF
Finding maximum diode forward current
I F =I out(max )× ( 1−Dmin )
I F =4 A × ( 1−0.5 )=2 A
Peak inverse voltage of a diode
PIV =V C=V out=8V
Finding drain source voltage of the Mosfet
When the Mosfet is On, Rds<1Ω
V SD ≈ 0.2 V

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QUESTION 6.
The linear series regulator in the circuit has following specifications.
Required Vout: 10V
Nominal Vin: 15 V
Nominal output current: 15 Ma
Q2 has a current gain : >100
Q1 has a current gain : 50.
a. If ZD1 is a 5.2V Zener diode , calculating suitable component values for R1,R2, R3 to meet
the requirement.
Finding R3 and R2.
V R 3=V out × R3
{ R2 + R3 } (i)
V R 3= 10 R3
{ R2 + R3 }
Also using Base-Emitter voltage and zener diode voltage
V R 3=V BE 2 +V ZD1 ( ii)
V R 3=0.7+ 5.2V =5.9V
Equating equation (i) and (ii)
5.9 V = 10 R3
{ R2 + R3 }
0.59 { R2 + R3 }=R3
0.41 R3=R2
If we let R3=100 kΩ, then R2=41 kΩ
Finding R1
V ¿=V R 1=15 V
R1= V R 1
I R 1
(iii)
Also
I R 1=I B 1+ IC 2 (iv)
Assuming I C 1 ≈ I E 1=Iout=15 mA

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I B 1= 15 mA
Q 1 = 15 mA
50 =0.3 mA
Since Q 2>100 ,then I B 2 ≈ 0 and therefore, using current loop R3, ZD1 and Q2,
I R 2=I R 3=IVDZ 1=I E 2 ≈ Iout=15 mA
Assuming IB 2 ≈ 0 then I E 2 =I C 2=15 mA
Replacing in (iv)
IR 1=15 mA +0.3 mA=15.3 mA
Substituting in equation (iii)
R1= 15 V
0.0153 A =980 Ω
Therefore, R1=980Ω , R2=41 kΩ∧R3=100 kΩ
b. Comparing efficiency of a shunt regulator to a series regulator at the same loas voltage
and current.
Shunt regulator is less efficient since it always draws current even when the system is not
subjected to the load unlike shunt regulator which does not draw current at no load
condition.
c. Replacing linear regulator with switching regulator.
ii) Advantages and disadvantages of replacing the linear regulator with a switching
regulator.
Advantages.
I. The efficiency of switching regulators is higher than linear regulators.
II. Switching regulators can step up or step down voltages with much ease as
compared to linear regulators
Disadvantage